Mixing
Binary Entropy Function (properties)
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I'm not a mathematician and I'm trying to understand the properties of the binary entropy function.
In particular, I would like to ask you, if the distributive law can be applied to the following expression: $(x_i - y_i) H(frac{1}{2^{i-1}}) quad s.t. quad x_i H(frac{1}{2^{i-1}}) - y_i H(frac{1}{2^{i-1}})$, without any violation of the $H$'s properties? Thank you in advance!
entropy
asked Feb 2 at 10:31
YiotaYiota
11
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add a comment |
$begingroup$
I'm not a mathematician and I'm trying to understand the properties of the binary entropy function.
In particular, I would like to ask you, if the distributive law can be applied to the following expression: $(x_i - y_i) H(frac{1}{2^{i-1}}) quad s.t. quad x_i H(frac{1}{2^{i-1}}) - y_i H(frac{1}{2^{i-1}})$, without any violation of the $H$'s properties? Thank you in advance!
entropy
asked Feb 2 at 10:31
YiotaYiota
11
$endgroup$
$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
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– Lord Shark the Unknown
Feb 2 at 10:36
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Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43
add a comment |
$begingroup$
I'm not a mathematician and I'm trying to understand the properties of the binary entropy function.
In particular, I would like to ask you, if the distributive law can be applied to the following expression: $(x_i - y_i) H(frac{1}{2^{i-1}}) quad s.t. quad x_i H(frac{1}{2^{i-1}}) - y_i H(frac{1}{2^{i-1}})$, without any violation of the $H$'s properties? Thank you in advance!
entropy
asked Feb 2 at 10:31
YiotaYiota
11
$endgroup$
I'm not a mathematician and I'm trying to understand the properties of the binary entropy function.
In particular, I would like to ask you, if the distributive law can be applied to the following expression: $(x_i - y_i) H(frac{1}{2^{i-1}}) quad s.t. quad x_i H(frac{1}{2^{i-1}}) - y_i H(frac{1}{2^{i-1}})$, without any violation of the $H$'s properties? Thank you in advance!
entropy
entropy
asked Feb 2 at 10:31
YiotaYiota
11
asked Feb 2 at 10:31
YiotaYiota
11
edited Feb 2 at 11:40
Yiota
asked Feb 2 at 10:31
YiotaYiota
11
asked Feb 2 at 10:31
YiotaYiota
11
asked Feb 2 at 10:31
YiotaYiota
11
11
$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 10:36
$begingroup$
Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43
add a comment |
$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 10:36
$begingroup$
Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43
$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 10:36
$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 10:36
$begingroup$
Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43
$begingroup$
Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43
add a comment |
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$begingroup$
Yes; this is the distributive law. It's just $(x-y)h=xh-yh$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 10:36
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Thank you for your prompt responses. So, if I understand well, the above expression does not violate any of the properties of the binary entropy function H and the distributive law is applied without any problem.
$endgroup$
– Yiota
Feb 2 at 10:43